Consider the LPP: Maximize $z = 5x + 3y$ subject to $3x+5y≤ 15,5x + 2y≤ 10,x,y ≥0$. The optimal feasible solution occurs at

$(\frac{20}{19},\frac{45}{19})$ only

The correct answer is Option (4) → $(\frac{20}{19},\frac{45}{19})$ only

Constraints: $3x+5y\le 15,\; 5x+2y\le 10,\; x,y\ge 0$

Corner points: $(0,0),\ (2,0),\ (0,3)$ and the intersection of $3x+5y=15$ with $5x+2y=10$.

Solve: $\begin{cases} 3x+5y=15\\ 5x+2y=10 \end{cases} \Rightarrow y=\frac{45}{19},\ x=\frac{20}{19}$

Objective values: $z(0,0)=0,\ z(2,0)=10,\ z(0,3)=9,\ z\!\left(\frac{20}{19},\frac{45}{19}\right)=\frac{235}{19}$.

Optimal point: $\left(\frac{20}{19},\,\frac{45}{19}\right)$